How many red nodes does a red-black-tree with black height bh(t) have(at most)?

Question

bh(t) = It is the number of black nodes on any simple path

A red black tree in our lecture is a binary search tree with

All node have 2 children(except leaf nodes)
Evry node is red or black
The root and all leaf nodes are black
All children of a red node are black
A Black node can at most have one red children
All paths from the root to the leaf have the same number of black nodes

I cant find the answer. Can someone help me please?

score 1 · Accepted Answer · answered Aug 21 '20 at 15:41

I do note that this definition is a bit more constrained than the usual one: in the standard definition requirement 5 is not present, and red nodes can be siblings. But given this extra restriction we can make the following analysis:

The only tree that has a black-height of 0 is the empty tree.
The only tree that has a black-height of 1 is a single black node (root & leaf)

The first tree that comes to mind that has a black-height of 2 is the following one:

                 b
                / \
               b   b

We can choose one of those two leaves to expand into a red node with two black children:

We cannot do more. There is no way to inject a second red node. The mirror of this tree is the only other possibility to stay within the constraints you have listed, and have a black-height of 2.

This shape is a key in incrementing to the next black-height of 3 (and greater). I will call this tree T.

To produce a tree with a black-height of 3, with as many red nodes as possible, we can replace each leaf (there are 3) with a copy of T. The result is this:

                 _____ b_____
                /            \
               R              b
             /   \           / \
           b       b        R   b
          / \     / \      / \
         R   b   R   b    b   b
        / \     / \
       b   b   b   b

Again, we can mirror the different subtrees, but there is no way to get more red nodes in a tree with a black-height of 3.

Now consider this: in this expansion we replaced each existing leaf with T, and since T has one red node, we actually added the same number of red nodes as there were leaves. And since T has 3 leaves, we multiplied the number of leaves by 3, which will determine how many T trees we will inject at the next expansion... etc.

This gives us a recurrence relation -- I write r_h to represent the maximum number of red nodes in a tree of black height h:

r₀ = 0
r₁ = 0
r₂ = 1
r_h = 1 + 3r_h-1

In a direct formula that last equation becomes:

r_h = ∑_{_i=1..h-2} 3ⁱ

Such sums can be rewritten as a repetition of 1 digits in base 3 notation, and so we can derive:

r_h = (3^h-1 - 1) / 2

This formula gives the correct result also for h=1 and h=2, so we only need to add the case for h=0:

r₀ = 0
r_h = (3^h-1 - 1) / 2, when h > 0

Here is a table with the first few results:

black height (h) | max number of red nodes (r[h])
-----------------+----------------------------------------
           0     |     0
           1     |     0
           2     |     1
           3     |     4
           4     |    13
           5     |    40
           6     |   121
           7     |   364
           8     |  1093
          ...    |   ...
           h     |  (3^(h-1) - 1) / 2

How many red nodes does a red-black-tree with black height bh(t) have(at most)?

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